Days Until Poisoned Plants Stabilize

A rectangular binary grid represents plants: 1 is poisoned and 0 is healthy. At the end of each day, every healthy plant with at least k poisoned neighbors becomes poisoned. The eight horizontally, vertically, and diagonally adjacent cells are neighbors.
All changes for a day are simultaneous. Return the number of days until a full day would produce no additional poisoned plants.

Function
daysUntilPlantsStabilize(grid: int[][], k: int) → int

Examples
Example 1
grid = [[1,0,0],[0,0,0],[0,0,0]]
k = 1
return = 2
The cells adjacent to the initial poison change on day 1, and the final distant cell changes on day 2.
Example 2
grid = [[1,0],[0,0]]
k = 2
return = 0
No healthy plant initially has two poisoned neighbors, so the grid is already stable.

Constraints
1 <= grid.length, grid[i].length <= 500.
Every cell is 0 or 1, and every row has the same length.
1 <= k <= 8.

The trick is multi-source BFS with neighbor counts. Keep a count array where each cell tracks how many of its 8 neighbors are poisoned. Push every initial poisoned cell into a queue. Process the queue one day at a time: for each cell in the current level, increment the count of its healthy neighbors. If a neighbor's count reaches k, mark it poisoned for the next level and add it to the next queue. Because all changes are simultaneous, you must not poison a cell mid-day. Collect new cells first, then mark them. The common pitfall is brute force, rescanning all 250,000 cells each day, which can take many days. Another pitfall is counting the final empty day. Return the number of levels that actually produced new poisoned cells. Each cell gets enqueued once, so the total work is O(m*n*8). StealthCoder is the hedge if the level-by-level bookkeeping slips under pressure on the live OA.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.